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I formulated a statement, which is hopefully true (at least I'm not knowledgeable enough to see a reason for it not to be). However, I'm struggling to come up with a proof.

Let $W_i \in \mathbb{R}^{d_i \times d_{i - 1}}, \ i = 1, 2, \ d := \min d_i$. Consider $R := W_2 W_1$ with $\textrm{rank}(R) \leq d$.

Can we obtain any matrix close to $R$ and of rank not larger than that of $R$ by small perturbations of the factors $W_j$?

More formally, let $\epsilon>0$ be given. Can we then find a $\delta>0$ such that for any $P$ with $\|P-R\|<\delta$ and $\textrm{rank}(P) \leq d$, there are $V_i$ such that $V_2V_1=P$ and $\|V_i-W_i\|<\epsilon$?

Does the statement seem correct to you? How could I prove it? Thank you.

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    $\begingroup$ The definition of perturbation is not clear at all (unless you use non-standard matrices, but I guess that you do not.) $\endgroup$
    – Fedor Petrov
    Commented May 7, 2017 at 14:29
  • $\begingroup$ @FedorPetrov, I'm sorry for a possible confusion. Dr. Remling is correct, I would like all $\tilde{W}_j$ to be "close" to $W_j$. What would be the correct way to state this? $\endgroup$
    – Serghei
    Commented May 7, 2017 at 18:26
  • $\begingroup$ @ChristianRemling, I do agree that the original statement could have been formulated for just two factors. However, I meant "how should a formulate what a perturbation means". $\endgroup$
    – Serghei
    Commented May 8, 2017 at 13:47
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    $\begingroup$ It's not identical, but see mathoverflow.net/questions/264606/… $\endgroup$
    – Noah Stein
    Commented May 9, 2017 at 0:54
  • $\begingroup$ @ChristianRemling, thank you, it does seem more coherent. $\endgroup$
    – Serghei
    Commented May 9, 2017 at 11:11

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